A049629 -id:A049629 - cp's OEIS frontend

A004292 Expansion of (1+x)^2/(1-18*x+x^2).

1, 20, 360, 6460, 115920, 2080100, 37325880, 669785740, 12018817440, 215668928180, 3870021889800, 69444725088220, 1246135029698160, 22360985809478660, 401251609540917720, 7200167985927040300

Offset: 0

N. J. A. Sloane

J. M. Alonso, Growth functions of amalgams, in Alperin, ed., Arboreal Group Theory, Springer, pp. 1-34, esp. p. 32.
P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 160, middle display.

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Hacène Belbachir, Soumeya Merwa Tebtoub, László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Pairwise sums of A049629.

f:= gfun:-rectoproc({a(n)=18*a(n-1)-a(n-2),a(0)=1,a(1)=20,a(2)=360},a(n),remember):
map(f, [$0..20]); # Robert Israel, Jun 01 2015

CoefficientList[Series[(1+x)^2/(1-18*x+x^2),{x,0,20}],x] (* Vincenzo Librandi, Jun 13 2012 *)
a[n_]:=1/2(1-(-1)^2^n+(20+9 Sqrt[5])((9+4 Sqrt[5])^(2 n)-1)/(9+4 Sqrt[5])^(n+1));Table[a[n] // FullSimplify,{n,0,20}] (* Gerry Martens, May 30 2015 *)

Vec((1+x)^2/(1-18*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = (1/2)*(1 - (-1)^2^n + (20+9*sqrt(5))*((9+4*sqrt(5))^(2*n) - 1)/(9+4*sqrt(5))^(n+1)). - Gerry Martens, May 30 2015

A085348 Ratio-determined insertion sequence I(0.264) (see the link below).

Original entry on oeis.org

1, 4, 19, 72, 341, 1292, 6119, 23184, 109801, 416020, 1970299, 7465176, 35355581, 133957148, 634430159, 2403763488, 11384387281, 43133785636, 204284540899, 774004377960, 3665737348901, 13888945017644, 65778987739319

Offset: 0

John W. Layman, Jun 24 2003

nonn

This is one of the "twin" ratio-determined insertion sequences (RDIS) that are "children" in the next generation below the "parent" sequences I(0.25024) (A004253) and I(0.26816) (A001353) in the recurrence tree of RDIS sequences. The RDIS twin of this sequence is A085349. See the link for an explanation of RDIS twin. See A082630 or A082981 for other recent examples of RDIS sequences.

Assuming that a(n) = 18a(n-2) - a(n-4) is true: For n >= 2, a(n) = (t(i+2n+2) - t(i))/(t(i+n+2) + t(i+n)*(-1)^(n-1)), where (t) is any recurrence of the form (4,1) without regard to initial values. With an additional initional 0 is this sequence the union of A060645 for even n and A049629 for odd n. - Klaus Purath, Sep 22 2024

John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [Broken link]
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [local copy, corrected]
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006 [Local copy]

Cf. A001353, A004253, A082630, A082981, A085349.

It appears that a(n)=18a(n-2)-a(n-4).
Apparently a(n)a(n+3) = -4 + a(n+1)a(n+2). - Ralf Stephan, May 29 2004
From Klaus Purath, Sep 22 2024: (Start)
Assuming that a(n) = 18a(n-2) - a(n-4) is true:
a(2n) = 5a(2n-1) - a(2n-2), n >= 1.
a(2n+1) = 4a(2n) - a(2n-1), n >= 1. (End)

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A305315 a(n) = sqrt(5b(n)^2 - 4), with b(n) = A134493(n) = Fibonacci(6n+1), n >= 0.

Original entry on oeis.org

1, 29, 521, 9349, 167761, 3010349, 54018521, 969323029, 17393796001, 312119004989, 5600748293801, 100501350283429, 1803423556807921, 32361122672259149, 580696784543856761, 10420180999117162549, 186982561199565069121, 3355265920593054081629, 60207804009475408400201, 1080385206249964297121989

Offset: 0

Wolfdieter Lang, Jul 10 2018

This sequence gives all solutions of one of two classes of positive proper solutions a(n) = a1(n) of the Pell equation a(n)^2 - 5*b(n) = -4 with b(n) = b1(n) = Fibonacci(6*n+1) = A134493(n). These solutions are obtained from the fundamental positive solution [1, 1] (to be read as a column vector) by application of positive powers of the automorphic matrix A = matrix([9, 20], [4, 9]) with determinant +1.
The other class of positive proper solutions is obtained similarly from the fundamental solution [11,5] and is given by [a5(n), b5(n)], with a5(n) = A305316(n) and b5(n) = A134497(n) = F(6*n+5), with the Fibonacci numbers F = A000045.
The remaining positive solutions are improper and are obtained by application of the same matrix A on the fundamental improper solution [4, 2]. They are given by [a3(n), b3(n)], with a3(n) = 4*A049629(n) and b3(n) = A134495(n) = 2*A007805(n).
Via the Cayley-Hamilton theorem the powers of the automorphic matrix A are: A^n = matrix([S(n) - 9*S(n-1), 20*S(n-1)], [4*S(n-1), S(n) - 9*S(n-1)]) with the Chebyshev polynomials S(n-1) = S(n-1, x=18) = A049660(n), n >= 0.
This shows that ordered Markoff (Markov) triples [1, y, m], with 1 <= y <= m, have for m  from the union of sets {m1(k)}{k>=0} U {m5(k)}{k>=0) U {m3(k)}_{k>=0)}, with mj(k) = F(6*k+j), for j = 1, 5, and 3, the unique solutions yj(k) = (3*F(6*k+j) - aj(k))/2  < mj(k), namely y1(k) = F(6*k-1) = A134497(k-1) with F(-1) = 1,  y5(k) = F(6*k+3) = A134495(k) and y3(k) = F(6*k+1) = A134493. The solutions with the + sign are excluded because they are > mj(k). This trisection of the odd-indexed Fibonacci numbers as m numbers shows again the well known fact that each of them appears as largest member in a Markoff triple if the smallest member is x = 1. The positions of the odd-indexed Fibonacci numbers in the Markoff sequence A002559 are given in A158381. The conjecture in this case is that the odd-indexed Fibonacci numbers appear as largest numbers only in the ordered Markov triples with x = 1. See, e.g., the Aigner reference for the general Frobenius-Markoff conjecture.
Also Lucas numbers that are congruent to 1 mod 4. - Fred Patrick Doty, Aug 03 2020

			The solutions of the first class of positive proper solutions [a1(n), b1(n)] of the Pell equation  a^2 - 5*b^2 = -4  begin: [1, 1], [29, 13], [521, 233], [9349, 4181], [167761, 75025], [3010349, 1346269], [54018521, 24157817], ...
The solutions of the second class of positive proper solutions [a5(n), b5(n)] begin: [11, 5], [199, 89], [3571, 1597], [64079, 28657], [1149851, 514229], [20633239, 9227465], [370248451, 165580141], ...
The solutions of the class of improper positive solutions [a3(n), b3(n)] begin: [4, 2], [76, 34], [1364, 610], [24476, 10946], [439204, 196418], [7881196, 3524578], [141422324, 63245986], ...

Aigner, Martin. Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013.

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000045, A002559, A007805, A049629, A049660, A134493, A134495, A134497, A158381, A305316.

Select[LinearRecurrence[{1, 1}, {1, 3}, 115], Mod[#, 4] == 1 &] (* Fred Patrick Doty, Aug 03 2020 *)

my(x='x+O('x^20)); Vec((1+11*x)/(1-18*x+x^2)) \\ Altug Alkan, Jul 11 2018

a(n) = sqrt(5*(F(6*n+1))^2 - 4), with F(6*n+1) = A134493(n), n >= 0.
a(n) = S(n, 18) + 11*S(n-1, 18), n >= 0, with the Chebyshev polynomials S(n, 18) = A049660(n+1) and S(-1, 18) = 0.
a(n) = 18*a(n-1) - a(n-2), n >= 1, with a(0)=1 and a(-1) = -11.
G.f.: (1 + 11*x)/(1 - 18*x + x^2).
a(n) = 2*sinh((6*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022

A305316 a(n) = sqrt(5b(n)^2 - 4) with b(n) = Fibonacci(6n+5) = A134497(n).

Original entry on oeis.org

11, 199, 3571, 64079, 1149851, 20633239, 370248451, 6643838879, 119218851371, 2139295485799, 38388099893011, 688846502588399, 12360848946698171, 221806434537978679, 3980154972736918051, 71420983074726546239, 1281597540372340914251, 22997334743627409910279, 412670427844921037470771, 7405070366464951264563599, 132878596168524201724674011

Offset: 0

Wolfdieter Lang, Jul 10 2018

This sequence gives all solutions of one of two classes of positive proper solutions a(n) = a5(n) of the Pell equation a(n)^2 - 5*b(n) = -4 with b(n) = b5(n) = F(6*n+5) = A134497(n), with the Fibonacci numbers F = A000045. These solutions are obtained from the fundamental positive solution [11, 5] (to be read as a column vector) by application of positive powers of the automorphic matrix A = matrix([9, 20], [4, 9]) of determinant +1.
The other class of positive proper solutions is obtained similarly from the fundamental solution [1,1] and is given by [a1(n), b1(n)], with a1(n) = A305315(n) and b1(n) = A134493(n) = F(6*n+1).
The remaining positive solutions are improper and are obtained by application of positive powers of the same matrix A on the fundamental improper solution [4, 2]. They are given by [a3(n), b3(n)], with a3(n) = F(6*n+3) = 4* A049629(n) and b3(n) = A134495(n) = 2*A007805(n).
For the explicit form of powers of the automorphic matrix A in terms of Chebyshev polynomials S(n, 18) see a comment in A305315.
The relation to a proof using this Pell equation of the well known fact that each odd-indexed Fibonacci number appears as largest member in Markoff (Markov) triples with smallest member 1 see also A305315.

			See A305315 for the three classes of solutions of this Pell equation

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000045, A007805, A049629, A049660, A134493, A134495, A134497, A305315.

I:=[11, 199]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Jul 22 2018

f[n_] := Sqrt[5 Fibonacci[6 n + 5]^2 - 4]; Array[f, 17, 0] (* or *)
CoefficientList[ Series[(x + 11)/(x^2 - 18x + 1), {x, 0, 18}], x] (* or *)
LinearRecurrence[{18, -1}, {11, 199}, 18] (* Robert G. Wilson v, Jul 21 2018 *)

x='x+O('x^99); Vec((11+x)/(1-18*x+x^2)) \\ Altug Alkan, Jul 11 2018

a(n) = sqrt(5*(F(6*n+5))^2 - 4), with F(6*n+5) = A134497(n), n >= 0.
a(n) = 11*S(n, 18) + S(n-1, 18), n >= 0, with the Chebyshev polynomials S(n, 18) = A049660(n+1) and S(-1, 18) = 0.
a(n) = 18*a(n-1) - a(n-2), n >= 1, with a(-1) = -1 and a(0) = 11.
G.f.: (11 + x)/(1 - 18*x + x^2).

A307937 Numbers that can be written as the sum of four or more consecutive squares in more than one way.

Original entry on oeis.org

3655, 3740, 4510, 4760, 5244, 5434, 5915, 7230, 7574, 8415, 11055, 11900, 12524, 14905, 17484, 18879, 19005, 19855, 20449, 20510, 21790, 22806, 23681, 25580, 25585, 27230, 27420, 28985, 31395, 34224, 37114, 39606, 41685, 42419, 44919, 45435, 45955, 48026, 48139, 48225, 49015, 53941, 57164, 62006

Offset: 1

Robert Israel, May 06 2019

nonn

Numbers that are in A174071 in two or more ways.
The first number with more than two representations as a sum of four or more consecutive positive squares is 147441 = 18^2 + ... + 76^2 = 29^2 + ... + 77^2 = 85^2 + ... + 101^2.
If x = 2*A049629(n) and y = A007805(n) for n >= 1 (satisfying the Pell equation x^2 - 5*y^2 = -1), then the sequence contains 5*x^2+10 = Sum_{(5*y-3)/2 <= i <= (5*y+3)/2} i^2 = Sum_{x-2 <= i <= x+2} i^2 = 25*y^2 + 5.

			a(1) = 3655 is in the sequence because 3655 = 8^2 + ... + 22^2 = 25^2 + ... + 29^2.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A007805, A049629, A174071.

N:= 10^5: # to get all terms <= N
R:= 'R':
dups:= NULL:
for m from 4 while m*(m+1)*(2*m+1)/6 <= N do
   for k from 1 do
       v:= m*(6*k^2 + 6*k*m + 2*m^2 - 6*k - 3*m + 1)/6;
       if v > N then break fi;
       if assigned(R[v]) then
         dups:= dups, v;
       else
         R[v]:= [k, k+m-1];
       fi;
od od:
sort(convert({dups},list));

M = 10^5;
dups = {}; Clear[rQ]; rQ[_] = False;
For[m = 4, m(m+1)(2m+1)/6 <= M, m++, For[k = 1, True, k++, v = m(6k^2 + 6k m + 2m^2 - 6k - 3m + 1)/6; If[v > M, Break[]]; If[rQ[v], AppendTo[dups, v], rQ[v] = True]]];
dups // Sort (* Jean-François Alcover, May 07 2019, after Robert Israel *)

A143790 Integer Quotients of Lucas Numbers; a rectangular array by downward antidiagonals.

Original entry on oeis.org

1, 3, 1, 4, 6, 1, 7, 41, 19, 1, 11, 281, 341, 46, 1, 18, 1926, 6119, 2161, 124, 1, 29, 13201, 109801, 101521, 15251, 321, 1, 47, 90481, 1970299, 4769326, 1875749, 103361, 844, 1, 76, 620166, 35355581, 224056801, 230701876, 33281921, 711491, 2206, 1, 123

Offset: 1

Clark Kimberling, Sep 01 2008

Every integer-valued quotient of Lucas numbers (excluding L(0)=2) is in this array.
(Row 1) = A000032 except for a(0)
(Row 2) = A049685
(Row 3) = A049629
(Column 2) = A110391 except for initial terms

			Q(3,2)=L(9)/L(3)=76/4=19.

Cf. A000032.

Row 1: L(k)/L(1), where L(k)=A000032(k) = k-th Lucas number, for k>=1;

Row n: L(2nk-n)/L(n) for n>=2, k>=1.

cp's OEIS Frontend

A004292 Expansion of (1+x)^2/(1-18*x+x^2).

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A085348 Ratio-determined insertion sequence I(0.264) (see the link below).

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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A305315 a(n) = sqrt(5*b(n)^2 - 4), with b(n) = A134493(n) = Fibonacci(6*n+1), n >= 0.

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A305316 a(n) = sqrt(5*b(n)^2 - 4) with b(n) = Fibonacci(6*n+5) = A134497(n).

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Magma

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A307937 Numbers that can be written as the sum of four or more consecutive squares in more than one way.

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Maple

Mathematica

A143790 Integer Quotients of Lucas Numbers; a rectangular array by downward antidiagonals.

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Author

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Comments

Examples

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Formula

A305315 a(n) = sqrt(5b(n)^2 - 4), with b(n) = A134493(n) = Fibonacci(6n+1), n >= 0.

A305316 a(n) = sqrt(5b(n)^2 - 4) with b(n) = Fibonacci(6n+5) = A134497(n).